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October 2022 Jobs notifications 🔔 @allcoding1_official 🎯Infosys 9.5LPA 🎯TCS 🎯 Accenture 3.33LPA 🎯 Amazon 5LPA 🎯 cognizant 4.5 LPA 🎯 Capgemini 4LPA 🎯 Qualcomm 8LPA 🎯 Wipro 🎯 Zoho 4LPA 🎯Alter 6LPA 🎯IBM 5LPA 🎯HCL 4.25 LPA 🎯DXC 4LPA 🎯LTI 4LPA 🎯 KPMG 4LPA @allcoding1_official 🎯Zeta 3LPA 🎯Theyve 3.5LPA 🎯 Emerson 4LPA 🎯Hexaware 4LPA 🎯 Global Logic 4.2LPA 🎯 Google 7.5LPA 🎯 Pepsico 7LPA 🎯 Phonepe 5LPA 🎯 Sutherland 5.5LPA Guys♥️ Once check All notifications and apply Now👇👇👇👇 http://www.allcoding1.com INFOSYS Exam Ans (22/10/22) 3:00PM ↗️Share with your friends and groups Telegram:- @allcoding1

C++ Telegram:- @allcoding1

magic Number Code python 3 All test cases passed Telegram:- @allcoding1

XOR elements C++ language Telegram:- @allcoding1

XOR code do you want

C++ Question:- You are given Queries.each query consists of 3 integer X[i],Y[i] and N[i] where i donates the ith query The answer to ith query X[i],Y[i] and N[i] is denoted By (X[i],Y[i])^((X[i]+1),Y[i])^....^((X[i],N[i]*Y[i]) Find the sum of all answers modulo 10^9+7 Telegram:- @allcoding1

You are given Queries.each query consists of 3 integer X[i],Y[i] and N[i] where i donates the ith query The answer to ith query X[i],Y[i] and N[i] is denoted By (X[i],Y[i])^((X[i]+1),Y[i])^....^((X[i],N[i]*Y[i]) Find the sum of all answers modulo 10^9+7

#include <bits/stdc++.h> using namespace std; int count(string s) { int N, i, cnt = 0, ans = 0; N = s.length(); for (i = 0; i < N; i++) { if (s[i] == 'R') cnt++; if (s[i] == 'L') ans += cnt; } return ans; } int main() { string s = "RRLL"; cout << count(s) << endl; return 0; } C++ Telegram - @allcoding1

def getLargestString(s, k):     frequency_array = [0] * 26     for i in range(len(s)):         frequency_array[ord(s[i]) -                         ord('a')] += 1     ans = ""     i = 25     while i >= 0:         if (frequency_array[i] > k):             temp = k             st = chr( i + ord('a'))                          while (temp > 0):                 ans += st                 temp -= 1                        frequency_array[i] -= k             j = i - 1                          while (frequency_array[j] <= 0 and                    j >= 0):                 j -= 1             if (frequency_array[j] > 0 and                 j >= 0):                 str1 = chr(j + ord( 'a'))                 ans += str1                 frequency_array[j] -= 1                          else:                 break         elif (frequency_array[i] > 0):             temp = frequency_array[i]             frequency_array[i] -= temp             st = chr(i + ord('a'))             while (temp > 0):                 ans += st                 temp -= 1         else:             i -= 1                  return ans           if name == "main":        S = input()     k = 3     print (getLargestString(S, k)) Python Bob code Telegram:- @allcoding1

Python Find Minimum Distance Between two numbers in Array code Telegram:- @allcoding1

if (n > k && n > a) { printf("N is the Biggest number is %d", n); return n; } if (k > n && k > a) { printf("k is the Biggest number is %d", k); return k; } if (a > n && a > k) { printf("a is the Biggest number is %d", a); return a; } Telegram:- @allcoding1

from itertools import permutations def count(arr):     z=[]     perm = permutations(arr)          for i in list(perm):         z.append(list(i))     q=[]          for i in range(len(arr)-1):         x,y=arr[i],arr[i+1]                  for j in range(len(z)):             if z[j].index(x)!=len(z[j])-1:                 if z[j][z[j].index(x)+1]==y:                     q.append(z[j])                          for i in range(len(q)):          if q[i] in z:              z.remove(q[i])     return len(z) a= int(input()) b=list(map(int,input().strip().split())) print(count(b)) Python Telegram - @allcoding1

Java language Telegram:- @allcoding1

Java language Telegram:- @allcoding1